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Math Help - Image and Kernel check please

  1. #1
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    Image and Kernel check please

    T(x1,x2,x3,x4) = (4x1-x2, 5x2 + 3x4, -x3)

    Image = (4,0,0) (-1,5,0) (0,0,-1) (0,3,0)

    So the vectors that form a basis are (4,0,0) (0,0,-1) (0,3,0)

    Kernel

    4x1 - x2 = 0
    5x2 + 3x4 = 0
    -x3 = 0

    x1 = 1/4 x2
    x2 = x2
    x3 = 0
    x4 = -5/3 x2

    (x1, x2, x3, x4) = (1/4x2, x2, 0, -5/3 x2)

    = x2(1/4, 1, 0, -5/3)

    Kernel = (1/4, 1, 0, -5/3)
    Last edited by adam_leeds; August 28th 2010 at 10:58 AM.
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  2. #2
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    Im not sure if my basis for the image is right.
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  3. #3
    MHF Contributor

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    Yes, those are correct because (-1, 5, 0)= (-1/4)(4, 0, 0)+ (5/3)(0, 3, 0).

    (Actually, any two of (4, 0, 0), (0, 3, 0), and (-1, 5, 0), together with (0, 0, -1) would be a basis. In fact, just (1, 0, 0), (0, 1, 0), (0, 0, 1) is a basis because the image is all of R^3.)
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